A projectile is fired from horizontal ground with speed v and projection angle \theta . When the acceleration due to gravity is g , the range of the projectile is d . If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is g^\prime=\fracg0.81 , then the new range is d^\prime=n d . The value of n is \underline\hspace2cm .

Motion PYQ — JEE Advanced 2022 · Shift 1 · Q44 with Solution

View written solutionFree

Correct answer: 0.93TO0.97

Original range on uniform gravity $g$

For a projectile projected with speed $v$ at angle $\theta$ from horizontal ground,

$T=\frac{2v\sin\theta}{g}$

So the range is

=\frac{v^2\sin 2\theta}{g}.$$ --- 2. **Motion up to the highest point** At the highest point, vertical velocity becomes zero. Time to reach highest point: $$t_1=\frac{v\sin\theta}{g}.$$ Horizontal distance covered till highest point: $$x_1=v\cos\theta\cdot t_1=\frac{v^2\sin\theta\cos\theta}{g}= rac{d}{2}.$$ Maximum height: $$H=\frac{v^2\sin^2\theta}{2g}.$$ --- 3. **Entering new region at the highest point** After the highest point, effective gravity becomes $$g' = \frac{g}{0.81}.$$ At the highest point: - horizontal velocity remains $v\cos\theta$ - vertical velocity is $0$ - height above ground is $H$ Now the projectile falls from height $H$ under gravity $g'$ with initial vertical velocity zero. So the downward time is $$t_2=\sqrt{\frac{2H}{g'}}.$$ Substitute $H=\frac{v^2\sin^2\theta}{2g}$: $$t_2=\sqrt{\frac{2}{g'}\cdot \frac{v^2\sin^2\theta}{2g}} = v\sin\theta\sqrt{\frac{1}{gg'}}.$$ Using $g' = \frac{g}{0.81}$, $$gg' = g\cdot \frac{g}{0.81}=\frac{g^2}{0.81}$$ hence $$\sqrt{\frac{1}{gg'}}=\sqrt{\frac{0.81}{g^2}}=\frac{0.9}{g}.$$ Therefore, $$t_2=\frac{0.9v\sin\theta}{g}.$$ --- 4. **Horizontal distance after the highest point** This distance is $$x_2=v\cos\theta\cdot t_2 =v\cos\theta\cdot \frac{0.9v\sin\theta}{g} =0.9\frac{v^2\sin\theta\cos\theta}{g}.$$ But $$\frac{v^2\sin\theta\cos\theta}{g}=\frac{d}{2}.$$ So, $$x_2=0.9\cdot \frac{d}{2}=0.45d.$$ --- 5. **New range** The new total range is $$d' = x_1 + x_2 = \frac{d}{2}+0.45d=0.95d.$$ Thus, $$n=\frac{d'}{d}=0.95.$$ --- 6. **Comparison with stored answer** Stored correct answer: $0.93$ to $0.97$ Our derived answer is $$n=0.95,$$ which lies in the given range. So it agrees with the stored answer.

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View written solutionFree

Correct answer: 0.93TO0.97

Original range on uniform gravity $g$

For a projectile projected with speed $v$ at angle $\theta$ from horizontal ground,

$T=\frac{2v\sin\theta}{g}$

So the range is

=\frac{v^2\sin 2\theta}{g}.$$ --- 2. **Motion up to the highest point** At the highest point, vertical velocity becomes zero. Time to reach highest point: $$t_1=\frac{v\sin\theta}{g}.$$ Horizontal distance covered till highest point: $$x_1=v\cos\theta\cdot t_1=\frac{v^2\sin\theta\cos\theta}{g}= rac{d}{2}.$$ Maximum height: $$H=\frac{v^2\sin^2\theta}{2g}.$$ --- 3. **Entering new region at the highest point** After the highest point, effective gravity becomes $$g' = \frac{g}{0.81}.$$ At the highest point: - horizontal velocity remains $v\cos\theta$ - vertical velocity is $0$ - height above ground is $H$ Now the projectile falls from height $H$ under gravity $g'$ with initial vertical velocity zero. So the downward time is $$t_2=\sqrt{\frac{2H}{g'}}.$$ Substitute $H=\frac{v^2\sin^2\theta}{2g}$: $$t_2=\sqrt{\frac{2}{g'}\cdot \frac{v^2\sin^2\theta}{2g}} = v\sin\theta\sqrt{\frac{1}{gg'}}.$$ Using $g' = \frac{g}{0.81}$, $$gg' = g\cdot \frac{g}{0.81}=\frac{g^2}{0.81}$$ hence $$\sqrt{\frac{1}{gg'}}=\sqrt{\frac{0.81}{g^2}}=\frac{0.9}{g}.$$ Therefore, $$t_2=\frac{0.9v\sin\theta}{g}.$$ --- 4. **Horizontal distance after the highest point** This distance is $$x_2=v\cos\theta\cdot t_2 =v\cos\theta\cdot \frac{0.9v\sin\theta}{g} =0.9\frac{v^2\sin\theta\cos\theta}{g}.$$ But $$\frac{v^2\sin\theta\cos\theta}{g}=\frac{d}{2}.$$ So, $$x_2=0.9\cdot \frac{d}{2}=0.45d.$$ --- 5. **New range** The new total range is $$d' = x_1 + x_2 = \frac{d}{2}+0.45d=0.95d.$$ Thus, $$n=\frac{d'}{d}=0.95.$$ --- 6. **Comparison with stored answer** Stored correct answer: $0.93$ to $0.97$ Our derived answer is $$n=0.95,$$ which lies in the given range. So it agrees with the stored answer.

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